Metamath Proof Explorer


Theorem ruclem4

Description: Lemma for ruc . Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014)

Ref Expression
Hypotheses ruc.1 φ F :
ruc.2 φ D = x 2 , y 1 st x + 2 nd x 2 / m if m < y 1 st x m m + 2 nd x 2 2 nd x
ruc.4 C = 0 0 1 F
ruc.5 G = seq 0 D C
Assertion ruclem4 φ G 0 = 0 1

Proof

Step Hyp Ref Expression
1 ruc.1 φ F :
2 ruc.2 φ D = x 2 , y 1 st x + 2 nd x 2 / m if m < y 1 st x m m + 2 nd x 2 2 nd x
3 ruc.4 C = 0 0 1 F
4 ruc.5 G = seq 0 D C
5 4 fveq1i G 0 = seq 0 D C 0
6 0z 0
7 ffn F : F Fn
8 fnresdm F Fn F = F
9 1 7 8 3syl φ F = F
10 dfn2 = 0 0
11 10 reseq2i F = F 0 0
12 9 11 eqtr3di φ F = F 0 0
13 12 uneq2d φ 0 0 1 F = 0 0 1 F 0 0
14 3 13 eqtrid φ C = 0 0 1 F 0 0
15 14 fveq1d φ C 0 = 0 0 1 F 0 0 0
16 c0ex 0 V
17 16 a1i 0 V
18 opex 0 1 V
19 18 a1i 0 1 V
20 eqid 0 0 1 F 0 0 = 0 0 1 F 0 0
21 17 19 20 fvsnun1 0 0 1 F 0 0 0 = 0 1
22 21 mptru 0 0 1 F 0 0 0 = 0 1
23 15 22 eqtrdi φ C 0 = 0 1
24 6 23 seq1i φ seq 0 D C 0 = 0 1
25 5 24 eqtrid φ G 0 = 0 1